Let f ( x ) = a x 3 + b x 2 + c x + d, be a polynomial function, find relation between a,b,c,d such that it's roots are in an arithmetic/geometric progression. (separate relations)So for the arithmetic progression I took let α = x 2 and r be the ratio of the arithmetic progression.We have: x 1 = α − 2 r , x 2 = α , x 3 = α + 2 rTherefore: x 1 + x 2 + x 3 = − b a = 3 α x 1 2 + x 2 2 + x 3 2 = 9 α 2 − 2 c a → 4 r 2 = b 2 − 3 a c 3 a 2 x 1 x 2 x 3 = α ( α 2 − 4 r 2 ) = − d a and we get the final result 2 b 3 + 27 a 2 d − 9 a b c = 0.How should I take the ratio at the geometric progression for roots?I tried something like x 1 = α q , x 2 = α , x 3 = α qTo get x 1 x 2 x 3 = α 3 but it doesn't really work out..Note:I have to choose from this set of answers: (a) a 2 b = c 2 d (b) a 2 b 2 = c 2 d (c) a b 3 = c 3 d (d) a c 3 = b 3 d (e) a c = b d (f) a 3 c = b 3 d

Question

Let   f  (  x  )  =  a      x    3    +  b      x    2    +  c  x  +  d, be a polynomial function, find relation between a,b,c,d such that it&#039;s roots are in an arithmetic/geometric progression. (separate relations)So for the arithmetic progression I took let   α  =      x    2   and r be the ratio of the arithmetic progression.We have:      x    1    =  α  −  2  r  ,        x    2    =  α  ,        x    3    =  α  +  2  rTherefore:      x    1    +      x    2    +      x    3    =  −      b    a    =  3  α      x    1    2    +      x    2    2    +      x    3    2    =  9      α    2    −  2      c    a    →  4      r    2    =                    b        2            −      3      a      c              3              a        2                  x    1        x    2        x    3    =  α  (      α    2    −  4      r    2    )  =  −      d    a  and we get the final result   2      b    3    +  27      a    2    d  −  9  a  b  c  =  0.How should I take the ratio at the geometric progression for roots?I tried something like      x    1    =      α    q    ,        x    2    =  α  ,        x    3    =  α  qTo get       x    1        x    2        x    3    =      α    3   but it doesn&#039;t really work out..Note:I have to choose from this set of answers:  (a)         a    2    b  =      c    2    d    (b)         a    2        b    2    =      c    2    d    (c)     a      b    3    =      c    3    d  (d)     a      c    3    =      b    3    d    (e)     a  c  =  b  d    (f)         a    3    c  =      b    3    d

erlentzed · Accepted Answer

Step 1Using your notations  a      x    3    +  b      x    2    +  c  x  +  d  =  a  (  x  −      α    q    )  (  x  −  α  )  (  x  −  α  q  )Expand the rhs to get after simplifications  a      x    3    −            a      α              (                  q          2                +        q        +        1        )              q        x    2    +            a              α        2                    (                  q          2                +        q        +        1        )              q    x  −  a      α    3  Step 2Compare the coefficients to get  b  =  −            a      α              (                  q          2                +        q        +        1        )              q    c  =            a              α        2                    (                  q          2                +        q        +        1        )              q    d  =  −  a      α    3

Widersinnby7 · Accepted Answer

Step 1To sum up, a cubic has its roots in arithmetic progression if and only if the arithmetic mean of the roots is a root of the cubic ( so equals one of the roots).For the geometric progression, we could use the same trick, and say that the geometric mean of       x    1  ,       x    2  ,       x    3   is a root of P. Alternatively, to avoid cubic roots, one considers the equivalent statement that       x    1        x    2        x    3   must equal one of the       x    i    3  . So we set up the equation   Q  =  0 with roots       x    1    3  ,       x    2    3  ,       x    3    3   and impose the condition that       x    1        x    2        x    3   is a root.The equation for       x    i    3   can be obtained readily by eliminating x from the equalities   y  =      x    3  ,   a      x    3    +  b      x    2    +  c  x  +  d. We get a cubic equation for y      a    3        y    3    +  (  3      a    2    d  −  3  a  b  c  +      b    3    )      y    2    +  (  3  a      d    2    −  3  b  c  d  +      c    3    )  y  +      d    3    =  0and the condition is that   −      d    a   is a root of this equation.Step 2In general, given a polynomial P of degree n, one can get the condition on the coefficients so that for some ordering of the roots we have an algebraic condition   F  (      x    1    ,  …  ,      x    n    )  =  0. We take all the possible permutations of   F  (      x          σ      (      1      )        ,  …  ,      x          σ      (      n      )        ) of F. The condition is that the product of all these permutations is 0.If we want m conditions       F    1    =  ⋯  =      F    m    =  0, we set up   F  =  ∑      t    i        F    i  , where       t    1  ,   …      t    m   are variables, consider all the possible permutations of F. The condition is that the product of all these is 0, as a polynomial in       t    1  ,   …      t    m

Let f(x)=ax^3+bx^2+cx+d, be a polynomial function, find relation between a,b,c,d such that it's roots are in an arithmetic/geometric progression. (separate relations)

Answered question

Answer & Explanation

New Questions in Algebra II